Respuesta :
The distribution of the fuel consumption can be considered to have a normal distribution with mean μ=278 mpg and standard deviation σ= 5.2 mpg.
You have to determine within what range of fuel consumption you find 99.7% of the automobiles of the market.
First, let's draw a sketch under the normal distribution to visualize the lower and upper limits of the range to determine:
In the normal curve, the green shaded area between the values I named x₁ and x₂ represents 99.7% of the distribution. The remaining 0.3% is in the little tails found below x₁ and above x₂.
You can express this shaded area as a probability, where X represents the variable of interest:
[tex]P(x_1\leq X\leq x_2)=0.997[/tex]To determine this interval, you have to determine the difference between the accumulated probability up to x₂ and the accumulated probability up to x₁, so that:
[tex]P(X\leq x_2)-P(X\leq x_1)=0.997[/tex]The probability accumulated until the lowest limit of the interval, x₁, is represented by the left tail shown in the image, so that:
[tex]P(X\leq x_1)=0.0015[/tex]And the probability accumulated until the highest limit of the interval, x₂, is equal to the area under the curve until that value, that is the sum of 0.0015 and 0.997:
[tex]\begin{gathered} P(X\leq x_2)=0.0015+0.997 \\ P(X\leq x_2)=0.9985 \end{gathered}[/tex]Knowing the accumulated probabilities for both limits of the interval, you can use the standard normal distribution to determine them.
The standard normal distribution is defined as follows:
[tex]Z=\frac{X-\mu}{\sigma}N(0,1)[/tex]Under the standard normal distribution, both values will accumulate the same probability, using these values of probability you can determine the z values and then reverse the standardization to "translate" the said Z-values to values of the variable:
Value of x₁
Under the standard normal distribution:
[tex]P(Z\leq z_1)=0.0015[/tex]The value under the standard normal distribution that has an accumulated probability of 0.0015 is z₁=-2.97
Using the formula of the standard normal distribution we can translate this value to a value of the variable:
[tex]z_1=\frac{x_1-\mu}{\sigma}[/tex]Write the expression for x₁:
[tex]\begin{gathered} z_1\cdot\sigma=x_1-\mu \\ x_1=(z_1\cdot\sigma)+\mu \end{gathered}[/tex]For μ=278 and σ= 5.2 mpg:
[tex]\begin{gathered} x_1=(-2.97\cdot5.2)+278 \\ x_1=-15.4444+278 \\ x_1=262.556 \end{gathered}[/tex]The lower value of the interval is x₁=262.56 mpg
Value of x₂
Under the standard normal distribution, this value also accumulates the same probability as under the normal distribution, so that:
[tex]P(Z\leq z_2)=0.9985[/tex]Using the distribution entry, you can determine the corresponding Z-value.
The value under the standard normal distribution that accumulates 0.9985 of probability is z₂= 2.97
Use the standard normal distribution to determine the value of x₂:
[tex]\begin{gathered} z_2=\frac{x_2-\mu}{\sigma} \\ _{} \end{gathered}[/tex]Write the expression for x₂
[tex]\begin{gathered} z_2\cdot\sigma=x_2-\mu \\ x_2=(z_2\cdot\sigma)+\mu \end{gathered}[/tex]For μ=278 and σ= 5.2 mpg:
[tex]\begin{gathered} x_2=(2.97\cdot5.2)+278 \\ x_{2}=15.444+278 \\ x_{2}=293.444 \end{gathered}[/tex]The upper value of the interval is x₂=293.44 mpg
99.7% of the automobiles are expected to be between the fuel consumption values of 262.56mpg and 293.44 mpg, you can express this symbolically as:
[tex]P(262.56\leq X\leq293.44)=0.997[/tex]
